Question: A particle moves along the curve $y=x^4+2x$ so that the $x$ -coordinate is increasing at a constant rate of $2$ units per second. What is the rate of change (in units per second) of the particle's $y$ -coordinate when the particle is at the point $(0,0)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $8$ (Choice C) C $20$ (Choice D) D $4$
Background When working with motion along a curve, we should remember that this motion can also be represented by a position vector $(x,y)$. The main difference is that we're not given the equations for $x$ and $y$ in terms of time $t$, but only the relationship between $x$ and $y$ themselves. This, however, shouldn't prevent us from assuming the expressions for $x$ and $y$ in terms of $t$ exist. As always, $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$. Setting up the math We are given that $\dfrac{dx}{dt}=2$ for any value of $t$. We are asked for the rate of change of the particle's $y$ -coordinate when the particle is at the point $(0,0)$. In other words, we are asked for the value of $\dfrac{dy}{dt}$ at the point $(0,0)$. Finding $\dfrac{dy}{dt}$ $\dfrac{dy}{dt}=8x^3+4$ Finding $\dfrac{dy}{dt}$ at $(0,0)$ The expression for $\dfrac{dy}{dt}$ only depends on the particle's $x$ -coordinate, which in our case is ${x}={4}$ : $\begin{aligned} \dfrac{dy}{dt}&=8({0})^3+4 \\\\ &=4 \end{aligned}$ In conclusion, the rate of change of the particle's $y$ -coordinate when the particle is at the point $(0,0)$ is $4$ units per second.